Intro to Chinese Remainder Theorem

Prologue

Chinese Remainder Theorem (CRT) is also known as Sun zi’s Theorem. It first appear on the Chinese book called Sūnzǐ Suànjīng, literally The Mathematical Classic of Master Sun/Master Sun’s Mathematical Manual. Here’s the math question in that book.

今有物不知其數，三三數之餘二，五五數之餘三，七七數之餘二，問物幾何？

There is something, but we do not know its exact quantity. When divided by 3, the remainder is 2; when divided by 5, the remainder is 3; when divided by 7, the remainder is 2. What is the quantity?

To solve this, we need to know what the CRT is.

Statement

Suppose n1,n2,…,nkn_1,n_2,…,n_k are positive integers which are pairwise coprime (i.e., the greatest common devisor of each pair is 1). For any given integers a1,a2,…,aka_1,a_2,…,a_k, there exists an integer xx that satisfies the following system of congruences:

x≡a1( mod n1)x≡a2( mod n2)⋮x≡ak( mod nk)\begin{aligned} & x \equiv a_1\left(\bmod n_1\right) \\ & x \equiv a_2\left(\bmod n_2\right) \\ & \vdots \\ & x \equiv a_k\left(\bmod n_k\right)\end{aligned}

Moreover, the solution of xx is unique modulo NN, where

N=n1⋅n2⋅…⋅nkN=n_1\cdot n_2 \cdot\ldots\cdot n_k

How it works

  1. Find NN: Compute the product N=n1⋅n2⋅…⋅nkN=n_1\cdot n_2 \cdot\ldots\cdot n_k.
  2. Compute Partial Products: For each ii, compute Ni=NniN_i=\frac{N}{n_i}.
  3. Find Modular Inverse: For each ii, find the modular inverse MiM_i of NiN_i modulo nin_i. That is, find MiM_i such that Ni⋅Mi≡1(modn)iN_i \cdot M_i \equiv 1\pmod n_i.
  4. Construct the solution: The solution xx can be constructed as:
    x=∑i=1kai⋅Ni⋅Mi(modN)x=\sum_{i=1}^k a_i \cdot N_i \cdot M_i\pmod N

Example

Suppose we want to solve the following system of congruences:

x≡2( mod 3)x≡3( mod 5)x≡2( mod 7)\begin{aligned} x\equiv 2\left(\bmod 3\right) \\ x \equiv 3\left(\bmod 5\right) \\ x \equiv 2\left(\bmod 7\right) \end{aligned}

  1. Compute N=3⋅5⋅7=105N=3\cdot5\cdot7=105.

  2. Compute N1=1053=35,N2=1055=21,N3=1057=15N_1=\frac{105}{3}=35,\quad N_2=\frac{105}{5}=21,\quad N_3=\frac{105}{7}=15.

  3. Find modular inverses:

    35⋅M1≡1( mod 3),21⋅M2≡1( mod 5),15⋅M3≡1( mod 7)35 \cdot M_1 \equiv 1(\bmod 3),\quad 21 \cdot M_2 \equiv 1(\bmod 5),\quad 15 \cdot M_3 \equiv 1(\bmod 7).

    Solving these, we get M1=2,M2=1,M3=1M_1=2,\quad M_2=1,\quad M_3=1

  4. Construct the solution:

    x=[(2⋅35⋅2)+(3⋅21⋅1)+(2⋅15⋅1)]( mod 105)x=(140+63+30)( mod 105)x=233( mod 105)x=23( mod 105)\begin{aligned} & x=[(2 \cdot 35 \cdot 2)+(3 \cdot 21 \cdot 1)+(2 \cdot 15 \cdot 1)](\bmod 105) \\ & x=(140+63+30)(\bmod 105) \\ & x=233(\bmod 105) \\ & x=23(\bmod 105)\end{aligned}

So, x=23x=23 is a solution to the system of congruences.